Chapter 5 Projectile motion
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Transcript of Chapter 5 Projectile motion
![Page 1: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/1.jpg)
Chapter 5 Projectile motion
![Page 2: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/2.jpg)
Chapter 4: straight line motion
that was ONLY vertical or
ONLY horizontal motion
![Page 3: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/3.jpg)
Chapter 5: considers motion that follows a diagonal path
or a curved path
![Page 4: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/4.jpg)
When you throw a baseball,the trajectory is a curved path.
![Page 5: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/5.jpg)
We are going to separate the motion of a projectile into independent x and y motions
![Page 6: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/6.jpg)
The vertical motion is not affected by the horizontal motion.
And the horizontal motion is not affected by the vertical motion.
![Page 7: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/7.jpg)
Observe: a large ball bearing is dropped
at the same time as a second ball bearing is fired horizontally.
![Page 8: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/8.jpg)
What happened?
![Page 9: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/9.jpg)
Remember
adding 2 perpendicular vectors
horizontal and
vertical vectors.
![Page 10: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/10.jpg)
When we add perpendicular vectors we use Pythagorean theorem to find the resultant.
![Page 11: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/11.jpg)
Consider a vector B that is pointed at an angle q wrt horizontal direction.
![Page 12: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/12.jpg)
We are going to break vector B into 2 perpendicular vectors:
Bx and By
![Page 13: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/13.jpg)
![Page 14: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/14.jpg)
![Page 15: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/15.jpg)
if you ADD vectors Bx + By
you get vector B.
![Page 16: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/16.jpg)
![Page 17: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/17.jpg)
Graphically, we can say:
![Page 18: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/18.jpg)
Draw a rectangle with vector B as the diagonal.
the component vectors Bx and By
are the sides of the rectangle
![Page 19: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/19.jpg)
![Page 20: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/20.jpg)
![Page 21: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/21.jpg)
Application:
![Page 22: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/22.jpg)
A Boat in a river
![Page 23: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/23.jpg)
How can we describe the motion of a boat in a river?
![Page 24: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/24.jpg)
The motion is affected by the motor of the boat
and by the current of the river
![Page 25: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/25.jpg)
Imagine a river 120 meters wide with a current of 8 m/sec.
![Page 26: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/26.jpg)
Imagine a river 120 meters wide with a current of 8 m/sec.
![Page 27: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/27.jpg)
If a boat is placed in the river [motor is off] ,
how fast will the boat drift downstream?
![Page 28: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/28.jpg)
If the boat is drifting, the total speed of the boat just equals
the speed of the current.
![Page 29: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/29.jpg)
for a boat drifting with the current:
Vtotal = Vboat + Vcurrent
Vtotal = -0 + Vcurrent = 8 m/sec
![Page 30: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/30.jpg)
Now suppose this boat can travel at a constant 15 m/sec when the
motor is on .
![Page 31: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/31.jpg)
What is the total speed of the boat downstream when the motor is on?
![Page 32: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/32.jpg)
The boat is traveling in the same direction as the current.
![Page 33: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/33.jpg)
V total downstream = Vtotal = Vboat + Vcurrent
Vtotal = 15↓ + 8↓ = 23 m/sec ↓
![Page 34: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/34.jpg)
What is the total speed of the boat traveling upstream
[against the current] ?
![Page 35: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/35.jpg)
The boat and current now move in opposite directions
![Page 36: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/36.jpg)
Vtotal = Vboat +Vcurrent Vtotal = 15 ↓+ ( 8 ↑)
Vtotal = 7↓ m/sec
![Page 37: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/37.jpg)
Summary:
traveling downstream: Vboat + Vcurrent
Traveling upstream:Vboat + [-Vcurrent]
![Page 38: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/38.jpg)
Crossing the river.
![Page 39: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/39.jpg)
If there was no current, how many seconds needed for this boat to travel 120 meters
from A to B?
![Page 40: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/40.jpg)
![Page 41: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/41.jpg)
Velocity = distance time
so time = distance velocity
![Page 42: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/42.jpg)
time = distance velocity
time = 120 m 15 m/sec
time = 8 seconds
![Page 43: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/43.jpg)
But there if IS a current. what happens when you try to go straight
across the river from A to B?
![Page 44: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/44.jpg)
The boat will travel from A to C.
![Page 45: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/45.jpg)
Every second the boat travels ACROSS 15 meters
and AT THE SAME TIME
every secondthe boat will be pushed
DOWNSTREAm 8 meters by the current .
![Page 46: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/46.jpg)
Vboat = 15 and
Vcurrent = 8↓
These velocities are perpendicular
![Page 47: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/47.jpg)
The RESULTANT velocity of the boat is
![Page 48: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/48.jpg)
Vresultant2 = Vboat
2 + Vcurrent2
Vresultant2 = 152 + 82
Vresultant2 = 225+ 64 =289
Vresultant = 17 m/sec
![Page 49: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/49.jpg)
The boat still crosses the river in 8 seconds ,
but it lands downstream at point C not at point B.
![Page 50: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/50.jpg)
![Page 51: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/51.jpg)
How far downstream is point C?
![Page 52: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/52.jpg)
Since the boat travels for 8 seconds
the current pushes the boat for 8 seconds
Vcurrent = 8 m/sec
![Page 53: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/53.jpg)
Velocity = distance/timeso
Distance = Velocity• time
Distance = 8 m/sec • 8 sec
distance downstream = 64 meters
![Page 54: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/54.jpg)
What is the total distance the boat travels?
D2 = Dx2 + Dy
2
![Page 55: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/55.jpg)
D2 = 1202 + 642 D2 = 14400+4096
D2 = 18496
D = 136 meters
![Page 56: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/56.jpg)
![Page 57: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/57.jpg)
![Page 58: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/58.jpg)
The triangles are similar:
![Page 59: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/59.jpg)
REMEMBEREvery second
the boat travels 15 meter across in the x direction IT ALSO TRAVELS
8 meter in the y direction
![Page 60: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/60.jpg)
What if you want to travel from point A to point B? Can you do that?
![Page 61: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/61.jpg)
You can cross from A to B if you point the boat in the correct direction.
![Page 62: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/62.jpg)
![Page 63: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/63.jpg)
Remember: Two perpendicular vectors can be
added to produce a single resultant vector that is pointed in
a specific direction.
![Page 64: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/64.jpg)
![Page 65: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/65.jpg)
SIMILARLY
ANY vector at angle q can be broken into the sum of two perpendicular vectors:one vector only in x direction
and one vector only in y direction.
![Page 66: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/66.jpg)
The magnitude of the component vectors is given by
Vx = Vocosq
Vy = Vosinq
![Page 67: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/67.jpg)
If you want to travel from A to B, you must direct the boat so that the “y component” of the boat’s
velocity cancels the velocity of the current.
![Page 68: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/68.jpg)
![Page 69: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/69.jpg)
![Page 70: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/70.jpg)
Point the boat so that the component of the boat’s velocity “cancels” the river
![Page 71: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/71.jpg)
Choose VboatY so that it is equal and opposite
to the Vcurrent
![Page 72: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/72.jpg)
VboatX = Vboatcosq
VboatY = Vboatsinq
![Page 73: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/73.jpg)
How do we find angle , q the direction to point the boat?
![Page 74: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/74.jpg)
Use arcsin or arctan
![Page 75: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/75.jpg)
arcsin = sin-1
Arcsine means “ the angle whose sine is” :
Sin-1 [VboatY/Vboat] = q
![Page 76: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/76.jpg)
Arctan = Tan-1 Arctan means
“ the angle whose tangent is”
![Page 77: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/77.jpg)
Arctan = Tan-1 Arctan means
“ the angle whose tangent is”
tan-1[Vboaty/Vboatx] = q
![Page 78: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/78.jpg)
Remember
![Page 79: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/79.jpg)
![Page 80: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/80.jpg)
PROJECTILE MOTION
![Page 81: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/81.jpg)
Projectile motion:
A projectile that has horizontal motion has a parabolic trajectory
We can separate the trajectory into x motion and y motion.
![Page 82: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/82.jpg)
In the x direction:
constant velocity
Vx = constant
distance in x direction X = Vx • t
![Page 83: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/83.jpg)
In y direction: free fall = constant acceleration.
Velocity in y direction : V = Vo – g t
Distance in y directionY = Yo + Vot – ½ g t2
![Page 84: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/84.jpg)
The range of a projectile is the maximum horizontal distance.
![Page 85: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/85.jpg)
Range and maximum height depend on the initial elevation angle.
![Page 86: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/86.jpg)
If you throw a projectile straight up,
the range = 0 height is maximum.
0 degrees : the minimum range but the maximum height.
![Page 87: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/87.jpg)
The maximum range occursat elevation 45o
![Page 88: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/88.jpg)
And for complementary angles
40 and 50 degrees30 and 60 degrees15 and 75 degrees10 and 80 degrees
![Page 89: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/89.jpg)
The range is identical for complementary angles
BUT the larger elevation angle gives a greater maximum height.
![Page 90: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/90.jpg)
![Page 91: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/91.jpg)
Remember:
![Page 92: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/92.jpg)
For a horizontal launch: Vo = initial horizontal velocity
0 = initial vertical velocity
in x direction: velocity is constant
in y direction: acceleration is constant
![Page 93: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/93.jpg)
If one object is fired horizontally at the same time
as a second object is dropped from the same height,
which one hits the ground first?
![Page 94: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/94.jpg)
Horizontal launch:in x [horizontal] direction
velocity is constant Vx = Vo
acceleration = 0
range = Vo • t [t = total time ]
![Page 95: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/95.jpg)
Horizontal launch:
In y direction:projectile is free falling.
Voy = 0Acceleration = g = 10 m/sec2↓
V = gt ↓d = ½ gt2
![Page 96: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/96.jpg)
Projectile motion lab:
part 1
![Page 97: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/97.jpg)
![Page 98: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/98.jpg)
Part 1: determine the velocity Vo of the projectile.
![Page 99: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/99.jpg)
![Page 100: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/100.jpg)
Projectile: - fired horizontally from height h.- follows parabolic path - Range R is where projectile hits the floor.
![Page 101: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/101.jpg)
![Page 102: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/102.jpg)
Equations: in y direction
Voy = 0g = constant acceleration
distance h = ½ gt2
![Page 103: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/103.jpg)
Equations in x-direction
acceleration = 0[constant velocity]
V= VoRange R = Vo ▪ t
![Page 104: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/104.jpg)
Part 1: fire projectile horizontally.
Measure all distances in METERS.
Measure starting height , h.
Measure range R.
![Page 105: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/105.jpg)
Part I Calculations:
![Page 106: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/106.jpg)
distance h = ½ gt2
measure h [ in METERS]
use g = 10 m/sec2
![Page 107: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/107.jpg)
solve equation to find t [ in seconds ]
distance h = ½ gt2
h= 5 t 2
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Measure value for R, the range in x direction
in METERS
![Page 109: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/109.jpg)
Use equation: R = Vo t∙
use measured value of R and calculated value for T
![Page 110: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/110.jpg)
Example: A projectile is fired horizontally from a table that is 2.0 meters tall. The projectile strikes the ground 3.6 meters from the edge of the table.
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![Page 112: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/112.jpg)
Given: H = 2.0 metersR = 3.6 meters
![Page 113: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/113.jpg)
h = ½ g t2
R = Vo t
![Page 114: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/114.jpg)
h = ½ g t2
2.0 = ½ [10] t2
![Page 115: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/115.jpg)
2.0 = ½ [10] t2
2.0 = 5 t2
2/5 = 0.40 = t2
![Page 116: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/116.jpg)
t = 0.63 sec
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For equation: R = Vo tuse
R = 3.6 m and
t = 0.63 sec
![Page 118: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/118.jpg)
R = 3.6 m = Vo [.63 sec] Vo = 3.6 m
0.63 sec
Vo = 5.7 m/sec
![Page 119: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/119.jpg)
The height of a projectile at any time along the path can be calculated.
![Page 120: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/120.jpg)
First calculate the height if there was no gravity.
If that case, a projectile would follow a straight line path
![Page 121: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/121.jpg)
![Page 122: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/122.jpg)
the projectile is always a distance 5t2 below this line.
![Page 123: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/123.jpg)
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Y = voy t – ½ gt2 Y = voy t – 5t2
i
![Page 125: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/125.jpg)
summary
![Page 126: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/126.jpg)
Vectors have magnitude and direction
Scalars have only magnitude
![Page 127: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/127.jpg)
The resultant of 2 perpendicular vectors
is the diagonal of a rectangle that has the 2 vectors as the sides.
![Page 128: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/128.jpg)
The perpendicular components of a vector are independent
of each other.
![Page 129: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/129.jpg)
The motion of a boat in a stream is the sum of a constant velocity of a boat [x dir] and the
constant velocity of the stream [y dir]
![Page 130: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/130.jpg)
The path of a boat crossing a stream is diagonal
![Page 131: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/131.jpg)
The horizontal component of a projectile is constant,
like a ball rolling on a surface with zero friction.
Objects in motion remain in motion at constant speed.
![Page 132: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/132.jpg)
The vertical component of a projectile is same as for an object in free fall.
![Page 133: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/133.jpg)
The vertical motion of a horizontally fired projectile is the same as free fall.
![Page 134: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/134.jpg)
For a projectile fired at an angle, the projectile will be 5t2 below where it would be if there was
no gravity.
![Page 135: Chapter 5 Projectile motion](https://reader035.fdocuments.us/reader035/viewer/2022062304/56813313550346895d99d3bf/html5/thumbnails/135.jpg)